Metric on product of metric spaces Let $X =X_1 \times X_2$ be the product of two metric spaces:
$$
{\mathrm{(}}{X}{\mathrm{,}}{d}_{1}{\mathrm{)}}\hspace{0.33em}\text{and}\hspace{0.33em}{\mathrm{(}}{X}{\mathrm{,}}{d}_{2}{\mathrm{)}}
$$
Show that the following are metrics on $X$.


*

*$
{d}{\mathrm{(}}{x}{\mathrm{,}}{y}{\mathrm{)}}\mathrm{{=}}\sqrt{{d}_{1}{\mathrm{(}}{x}_{1}{\mathrm{,}}{y}_{1}{\mathrm{)}}^{2}\mathrm{{+}}{d}_{2}{\mathrm{(}}{x}_{2}{\mathrm{,}}{y}_{2}{\mathrm{)}}^{2}}
$

*$
\mathit{\rho}{\mathrm{(}}{x}{\mathrm{,}}{y}{\mathrm{)}}\mathrm{{=}}\max\left[{{d}_{1}\mathrm{(}{x}_{1}\mathrm{,}{y}_{1}\mathrm{),}{d}_{2}\mathrm{(}{x}_{2}\mathrm{,}{y}_{2}\mathrm{)}}\right]
$
 A: It helps to think of working in the real vectorspace $\mathbb{R}^2$ and think about different norms, in your case the 2-norm and the $\infty$-norm.
As you pointed out in the comments, you are having trouble with the triangle inequality. First, observe that
$$
d(x,y) = \bigg\| \left( \begin{array}{c} d_1(x_1,y_1) \\ d_2(x_2,y_2)\end{array}\right)\bigg\|_2 = 
\sqrt{ d_1(x_1,y_1)^2 + d_1(x_1,y_1)^2}.
$$
Let $z = (z_1,z_2) \in X$. Then, you can use monotonicity of the square root and $(\cdot)^2$, namely that for positive $a,b$ you get
$$
a \leq b \implies \sqrt{a} \leq \sqrt{b} \text{ and } a^2 \leq b^2.
$$
Concretely, you calculate
$$
d_1(x_1,y_1) \leq d_1(x_1, z_1) + d_1(z_1,y_1) \implies d_1(x_1,y_1)^2 \leq d_1(x_1, z_1)^2 + d_1(z_1,y_1)^2
$$
and similarly for $d_2$. Then you can use monotonicity of the square root to derive an expression like this
$$
d(x,y) \leq \dots \leq  \bigg\| \left( \begin{array}{c} d_1(x_1,z_1) + d_1(z_1, y_1) \\ d_2(x_2,z_2) + d_2(z_2,y_2\end{array}\right)\bigg\|_2 = 
\bigg\| \left( \begin{array}{c} d_1(x_1,z_1) \\ d_2(x_2,z_2)\end{array}\right) + \left(\begin{array}{c} d_1(z_1,y_1) \\ d_2(z_2,y_2)\end{array}\right)\bigg\|_2 
\leq \dots
$$
Hopefully, you can take it from here and finish the proof. A similar line of argument applies to your second problem.
